Lemma 16.8.4. If for every Situation 16.8.1 where R is a field PT holds, then PT holds in general.
Proving Popescu approximation reduces to algebras over a field
Proof. Assume PT holds for any Situation 16.8.1 where R is a field. Let R \to \Lambda be as in Situation 16.8.1 arbitrary. Note that R/I \to \Lambda /I\Lambda is another regular ring map of Noetherian rings, see More on Algebra, Lemma 15.41.3. Consider the set of ideals
We have to show that \mathcal{I} is empty. If this set is nonempty, then it contains a maximal element because R is Noetherian. Replacing R by R/I and \Lambda by \Lambda /I we obtain a situation where PT holds for R/I \to \Lambda /I\Lambda for any nonzero ideal of R. In particular, we see by applying Proposition 16.5.3 that R is a reduced ring.
Let A \to \Lambda be an R-algebra homomorphism with A of finite presentation. We have to find a factorization A \to B \to \Lambda with B smooth over R, see Algebra, Lemma 10.127.4.
Let S \subset R be the set of nonzerodivisors and consider the total ring of fractions Q = S^{-1}R of R. We know that Q = K_1 \times \ldots \times K_ n is a product of fields, see Algebra, Lemmas 10.25.4 and 10.31.6. By Lemma 16.8.2 and our assumption PT holds for the ring map S^{-1}R \to S^{-1}\Lambda . Hence we can find a factorization S^{-1}A \to B' \to S^{-1}\Lambda with B' smooth over S^{-1}R.
We apply Lemma 16.8.3 and find a factorization A \to B \to \Lambda such that some \pi \in S is elementary standard in B over R. After replacing A by B we may assume that \pi is elementary standard, hence strictly standard in A. We know that R/\pi ^8R \to \Lambda /\pi ^8\Lambda satisfies PT. Hence we can find a factorization R/\pi ^8 R \to A/\pi ^8A \to \bar C \to \Lambda /\pi ^8\Lambda with R/\pi ^8 R \to \bar C smooth. By Lemma 16.6.1 we can find an R-algebra map D \to \Lambda with D smooth over R and a factorization R/\pi ^4 R \to A/\pi ^4A \to D/\pi ^4D \to \Lambda /\pi ^4\Lambda . By Lemma 16.7.2 we can find A \to B \to \Lambda with B smooth over R which finishes the proof. \square
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